Here i am again Newtons second law

Click For Summary

Homework Help Overview

The discussion revolves around applications of Newton's second law in various scenarios, including inclined planes and force interactions in a rope system. Participants explore the conditions for equilibrium and resultant forces acting on objects in motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conditions under which the acceleration of blocks is zero, emphasizing the balance of forces. Others explore the forces acting on segments of a rope and question the relationships between these forces. Additionally, there is a focus on breaking down forces into components and determining angles for accurate calculations.

Discussion Status

The conversation includes attempts to clarify the setup of problems and the relationships between forces. Some participants provide guidance on drawing free body diagrams and resolving forces into components, while others express confusion about angles and force relationships. Multiple interpretations of the problems are being explored.

Contextual Notes

Participants mention specific values and conditions, such as the mass of blocks and the angles of applied forces, while also noting the absence of certain information needed for complete resolution. There are indications of homework constraints influencing the discussion.

badman
Messages
57
Reaction score
0
Consider another special case in which the inclined plane is vertical (\theta=\pi/2). In this case, for what value of m_1 would the acceleration of the two blocks be equal to zero?
Express your answer in terms of some or all of the variables m_2 and g.

a good point in the direction will help
 

Attachments

  • 46055_a.jpg
    46055_a.jpg
    5.2 KB · Views: 794
Physics news on Phys.org
Draw a Free Body Diagram for each box, and i recommend for the inclined box to have a system of coordinates at the angle theta with respect to the horizontal.
 
The acceleration of the two blocks be equal to zero when the weight of Block 1 is equals to the tension of string. Then, there is no net force acting on Block 1, hence Block 2, the velocity is uniform or equals to zero.
 
thanks guys. can you help me with a another problem.
Now imagine two points, Q and P, that divide the rope into segments L, M ,and R. View Figure The rope remains stationary. Assume that segment L exerts a force of magnitude F_LM on segment M. What is the magnitude F_RM of the force exerted by segment R on segment M?
Give your answer in terms of F_LM and constants such as g.

shouldn't the force acting on segment R be equal to the force segment M exerted on it but in opposite direction?

i inputed F_lm *G but it said i was missing a multiplicative factor?
 

Attachments

  • MFS_3l_10_b.jpg
    MFS_3l_10_b.jpg
    3.9 KB · Views: 884
badman said:
i inputed F_lm *G but it said i was missing a multiplicative factor?
Why did you multiply by "G"?
 
whoops i was thinking of another answer to another problem i had. sorry bout that
 
another problem

nvm, i had another question but i already solved it
 
Last edited:
heres another i can't seem to right

The diagram below shows a block of mass m=2.0\; \rm kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F_1 = 4.0\;{\rm N}, F_2 = 6.0\;{\rm N}, and F_3 = 8.0\;{\rm N} are applied to the block, initially at rest on the surface, at angles shown on the diagram. View Figure In this problem, you will determine the resultant (total) force vector from the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x-axis (i.e., all angles are positive).

this is the question
Calculate the magnitude of the total resultant force \vec{F}_{\rm r} = \vec{F}_1+ \vec{F}_2 +\vec{F}_3 acting on the mass.
Express the magnitude of the resultant force in Newtons to two significant figures.



i already know the formula and that i need to break this down into compenents. for each force vector.
the confusing thing is the angles, that i need.

this is basically how i set it up. Fx1= N*cos155
fy1=N*sin155
fx2 and fy2 i can't seem to get
fx3=N*cos325
fy3=Ncos325


are the angles right, i can't seem to figure out the angle for the vector F_2
 

Attachments

  • 1206.jpg
    1206.jpg
    4.5 KB · Views: 765
badman said:
this is basically how i set it up. Fx1= N*cos155
fy1=N*sin155
fx2 and fy2 i can't seem to get
fx3=N*cos325
fy3=Ncos325


are the angles right, i can't seem to figure out the angle for the vector F_2
The angles are not right. As marked in the diagram, the angle that each vector makes with the +x axis is:
F1 is at 25 degrees (not 155);
F2 is at 325 degrees;
F3 is at 180 degrees (it points in the -x direction, which is 180 degrees from the +x direction).
 
  • #10
thanks doc.
 

Similar threads

Replies
44
Views
4K
Replies
13
Views
4K
Replies
15
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
11K
Replies
20
Views
4K
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K